This invention relates to a differential amplifier.
A differential amplifier, as shown in FIG. 4, improved in non-linear transfer characteristic has been proposed in Japanese Unexamined Patent Application No. 224411/1990. As shown in FIG. 4, the emitters of NPN transistors 1 and 2 are connected to each other through a resistor 7, and connected to current sources 13 and 14, respectively, which are adapted to supply bias currents to the transistors. The collector current I2 of the transistor 1 is applied to the emitter of a transistor 3, while the collector current I1 of the transistor 2 is applied to the emitter of a transistor 4. The base of the transistor 1 is connected to the emitter of the transistor 4, while the base of the transistor 2 is connected to the emitter of the transistor 3. A signal source 11 provides an input signal Vi between the bases of the transistors 3 and 4. A voltage source 9 applies a bias voltage Vr to the bases of the transistors 3 and 4.
Another voltage source 10 supplies a supply voltage Vcc directly to the collector of the transistor 3, and through a load resistor 8 to the collector of the transistor 4. The differential amplifier thus organized provides an output signal Vout at the connecting point of the load resistor 8 and the collector of the transistor 4. The output signal is supplied to an output terminal 12.
In the differential amplifier, the collector currents I1 and I2 change in response to the input signal Vi. When the input signal Vi increases in a positive direction with the base of the transistor 3 as positive side, the collector current I1 increases whereas the collector current I2 decreases. Assuming that the base currents of the transistors 1 and 2 can be disregarded because they are much smaller than the collector currents I1 and I2, then in the case when the input signal Vi is 0 V, I1=I2=I0 where I0 is the output current of each of the current sources 13 and 14.
The input signal Vi can be expressed by the following Equation (1): EQU Vi=Vbe3+Vbe2+(I1-I0).multidot.R7-(Vbe1+Vbe4) . . . (1)
where R7 is the resistance of the resistor 7, and Vbe1, Vbe2, Vbe3 and Vbe4 are the base-emitter voltages of the transistors 1, 2, 3 and 4, respectively.
The relation between the base-emitter voltage Vbe1 and the current I2 can be represented by the following Equation (2) EQU I2=Is.multidot.{exp (Vbe1/Vt)-1} (2)
where Is is the saturation current of the transistor, and Vt is k.multidot.T/q in which k is Boltzmann's constant, T is the absolution temperature, and q is the electron charge.
By solving Equation (2) for Vbe1, the following Equation (3) can be obtained: EQU Vbe1=Vt.multidot.ln (I2/Is+1) (3)
In the case where the transistors 1 and 2 are equal in characteristics, then the base-emitter voltage Vbe2 can be represented by the following Equation (4): EQU Vbe2=Vt.multidot.ln (I1/Is+1) (4)
In the case where the transistors 3 and 4 have the same characteristic as the transistors 1 and 2 and the base currents of the transistors 3 and 4 are disregarded, the base-emitter voltages Vbe3 and Vbe4 of the transistors 3 and 4 are as follows: EQU Vbe3=Vt.multidot.ln (I2/Is+1) (5) EQU Vbe4=Vt.multidot.ln (I1/Is+1) (6)
Insertion of Equations (3), (4), (5) and (6) in Equation (1) results in the following Equation (7): ##EQU1##
By solving Equation (7) for I1, the following Equation (8) is obtained: EQU I1=I0+Vi/R7 (8)
Because I1+I2=2.multidot.I0, therefore, the following Equation (9)is obtained: EQU I2=I0-Vi/R7 (9)
When the resistance of the resistor 8 is RL, the output signal Vout can be expressed as follows: ##EQU2##
As is seen from Equation (10), Vout is obtained by amplifying Vi with a gain of RL/R7.
As is apparent from the above description, in the circuit shown in FIG. 4, the distortion component (which is the non-linearity attributing to the relation between the collector current and the base-emitter voltage of a transistor) is eliminated on the basis of the relation of Vi and Vout. That is, the differential amplifier is excellent in linearity.
On the other hand, in the above-described differential amplifier, the collector-base voltage Vcb1 of the transistor 1 is expressed as follow: ##EQU3##
Similarly, the collector-base voltage Vcb2 of the transistor 2 can be represented by the following Equation (12): EQU Vcb2=Vbe3-Vbe4-Vi (12)
In the case where the input signal is large in amplitude, the base-emitter voltages Vbe3 and Vbe4 can be regarded as substantially equal to each other, and therefore Equations (11) and (12) can be approximated as follows: EQU Vcb1.apprxeq.Vi (13) EQU Vcb2.apprxeq.-Vi (14)
As is seen from Equations (13) and (14), the collector-base voltages of the transistor 1 and 2 become negative in accordance with the input signal Vi, so that the transistors Operate in the saturation region. When a transistor operates in the active region, the current amplification factor hfe is greatly decreased, and the base current is accordingly increased. That is, in this case, unlike the above-described case, the base current cannot be disregarded. In Equation (7), the base-emitter voltages Vbe corresponding to the non-linear components are not included because Vbe1=Vbe3 and Vbe2=Vbe4 are satisfied. However, these relations are not satisfied in the case where the transistors 3 and 4 operate in the active region whereas the transistors 1 and 2 operate in the saturation region. Hence, when the input signal is large in amplitude, then Equation (7) is no longer established. Thus, the conventional differential amplifier suffers from the difficulty that it is small in input dynamic range.